Electron Emission

Dual Nature of Radiation and Matter · CBSE

Theory

The definition of work function, the three emission processes with one example each, and the background on how the electron was identified. One- and two-mark questions come from here; the 2023 MCQ on the energy of a photon of wavelength $\lambda$ was also filed under this section.

Passing electricity through a gas at low pressure — about $0.001$ mm of mercury — in a discharge tube produces a fluorescent glow on the glass opposite the cathode (yellowish-green for soda glass). The radiation causing it was traced to the cathode and named cathode rays.
William Crookes discovered cathode rays in 1870 and suggested in 1879 that they are streams of fast-moving negatively charged particles.
J. J. Thomson (1897) confirmed this. By applying mutually perpendicular electric and magnetic fields across the discharge tube he measured, for the first time, the speed and the specific charge of the cathode-ray particles:

$\dfrac{e}{m} = 1.76\times10^{11}\ \text{C kg}^{-1}, \qquad v \approx (0.1 \text{ to } 0.2)\,c$

Crucially, $e/m$ came out the same whatever the cathode material and whatever gas filled the tube. That universality is what made Thomson call the particles electrons and identify them as fundamental constituents of all matter. Nobel Prize, 1906.
In 1887 it was found that some metals irradiated with ultraviolet light emitted negatively charged particles, and that some metals heated strongly did the same. The $e/m$ of these particles matched that of cathode rays — the same particle, produced three different ways.
R. A. Millikan's oil-drop experiment (1913) found the charge on a drop always to be an integral multiple of $1.602\times10^{-19}$ C, establishing that electric charge is quantised. Combining $e$ with $e/m$ gives the electron mass, $m_e = 9.1\times10^{-31}$ kg.
Roentgen discovered X-rays in 1895; together with Thomson's electron these were the milestones that opened up atomic structure.

A metal contains free electrons, which is why it conducts. They move about inside the metal in what is, to a good approximation, a constant potential.
They are not free to leave. If an electron starts to come out, the surface it leaves behind acquires a positive charge and pulls the electron straight back. The attractive forces of the positive ions hold the electron inside the surface.
So an electron can escape only if it is given enough energy to overcome that pull.

Definition to write in the exam: the work function of a metal is the minimum energy required by an electron to escape from the metal surface.
Symbol $\phi_0$ (many CBSE keys write $W_0$ ); measured in electron volt.

$1\ \text{eV} = 1.602\times10^{-19}\ \text{J}$

One electron volt is the energy gained by an electron accelerated through a potential difference of one volt.
$\phi_0$ depends on the metal and on the nature (cleanliness, coating, oxidation) of its surface. It does not depend on the incident light in any way.
Free electrons in a metal do not all have the same energy — they have an energy distribution, and Pauli's exclusion principle makes it unlike the Maxwell distribution of a gas. So the energy needed to get out differs from electron to electron; the work function is the least such energy, for the most loosely bound electrons. This is exactly why photoelectrons emerge with a spread of kinetic energies.

Process	How the energy is supplied	Typical example
Thermionic emission	the metal is heated, so thermal energy is imparted to free electrons	heated cathode of a vacuum tube or CRT
Field emission	a very strong electric field, of the order of $10^{8}$ V m $^{-1}$ , pulls electrons out	spark plug
Photoelectric emission	light of suitable frequency falls on the metal	photocell, solar-cell type detectors

The electrons released in the third case are called photoelectrons. That process is the subject of the rest of the chapter.

$\phi_0 = h\nu_0 = \dfrac{hc}{\lambda_0}$

$\nu_0$ = threshold frequency, $\lambda_0$ = threshold wavelength. Given $\phi_0$ in eV, $\lambda_0 = 1240/\phi_0$ nm.

The energy of a photon of wavelength $\lambda$ is $hc/\lambda$ , not $hc\lambda$ , $\lambda/hc$ or $\lambda h/c$ .

$E = h\nu, \qquad c = \nu\lambda \;\Rightarrow\; \nu = \dfrac{c}{\lambda} \;\Rightarrow\; E = \dfrac{hc}{\lambda}$

Check the answer by units if you are unsure: $h$ is J s, $c$ is m s $^{-1}$ , $\lambda$ is m, so $hc/\lambda$ is joule.

$e/m = 1.76\times10^{11}$ C kg $^{-1}$ , $e = 1.6\times10^{-19}$ C, $m_e = 9.1\times10^{-31}$ kg.
Work function values CBSE has used: Cs $2.14$ eV, K $2.3$ eV, Na $2.75$ eV, Ca $3.20$ eV, Mo $4.17$ eV, Ni $5.15$ eV. (One 2018 paper quoted Na $1.92$ eV and K $2.15$ eV — always use the paper's own values.)
Worked check: for a work function of $4.14$ eV, $\lambda_0 = hc/\phi_0 = (6.6\times10^{-34}\times3\times10^{8})/(4.14\times1.6\times10^{-19}) = 3\times10^{-7}$ m $=3000$ Å.

Every numerical set on this subtopic. Full working is at the end of the page.

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.Answer: (B)↑ back to the question
Energy of a photon of wavelength $\lambda$ : $\;E=h\nu$ , and since $c=\nu\lambda$ , $\nu=c/\lambda$ , so $\;E=\dfrac{hc}\lambda$ . Option (b).